Question

What is the volume of 0.640 grams of Oz gas at Standard Temperature and Pressure (STP)?

Answer 1

Answer: The volume of 0.640 grams of $O_{2}$ gas at Standard Temperature and Pressure (STP) is 0.449 L.

Explanation:

Given: Mass of $O_{2}$ gas = 0.640 g

Pressure = 1.0 atm

Temperature = 273 K

As number of moles is the mass of substance divided by its molar mass.

So, moles of $O_{2}$ (molar mass = 32.0 g/mol) is as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{0.640 g}{32.0 g/mol}\\= 0.02 mol$

Now, ideal gas equation is used to calculate the volume as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

$PV = nRT\\1.0 atm \times V = 0.02 mol \times 0.0821 L atm/mol K \times 273 K\\V = 0.449 L$

Thus, we can conclude that the volume of 0.640 grams of $O_{2}$ gas at Standard Temperature and Pressure (STP) is 0.449 L.